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Ufumbuzi - Tiger algebra Suluhishaji

42817.15

42817.15

Tazama hatua Jifunze zaidi na Tiger Jaribu hili:

Maelezo kwa hatua

1. Weka fomula ya riba changamano

$c i (22822, 3, 12, 21)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 822$ , $r = 3 %$ , $n = 12$ , $t = 21$ .

$c i (22822, 3, 12, 21)$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 822$ , $r = 3 %$ , $n = 12$ , $t = 21$ .

$P = 22822, r = 3 %, n = 12, t = 21$

$\frac{3}{100} = 0.03$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 822$ , $r = 3 %$ , $n = 12$ , $t = 21$ .

$A = P {(1 + r / n)}^{(n t)}$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 822$ , $r = 3 %$ , $n = 12$ , $t = 21$ .

$N e x t c o m p u t e r / n, n t, a n d {(1 + r / n)}^{(n t)} .$

Tumia fomula $A = P {(1 + r / n)}^{(n t)}$ ukiwa na $P = 22, 822$ , $r = 3 %$ , $n = 12$ , $t = 21$ .

2. Hesabu kipengele cha ukuaji

$\frac{r}{n} = 0.0025$

$n t = 252$

${(1 + r / n)}^{(n t)}$

$r / n = 0.0025, n t = 252, {(1 + r / n)}^{(n t)} = 1.8761350008$ .

${(1 + r / n)}^{(n t)}$

${(1 + 0.0025)}^{252} = 1.8761350008$

$r / n = 0.0025, n t = 252, {(1 + r / n)}^{(n t)} = 1.8761350008$ .

$1.8761350008$

$r / n = 0.0025, n t = 252, {(1 + r / n)}^{(n t)} = 1.8761350008$ .

3. Hesabu kiasi cha mwisho

$A = 42817.15$

$42817.15$

$22, 822 \cdot 1.8761350008 = 42817.15$ .

$42817.15$

$22, 822 \cdot 1.8761350008 = 42817.15$ .

$42817.15$

$22, 822 \cdot 1.8761350008 = 42817.15$ .

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Jifunze zaidi na Tiger

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Tiger algebra Suluhishaji

Ufumbuzi - Tiger algebra Suluhishaji

Maelezo kwa hatua

1. Weka fomula ya riba changamano

2. Hesabu kipengele cha ukuaji

3. Hesabu kiasi cha mwisho

Kwa nini kujifunza hii

Vigezo na mada

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